Lecture 13: Zero-sum games using Experts

A clarification about today’s lecture:

The Oracle. My explanation for why the oracle was an easy problem was not very clear. Let me try again. Let us define

\displaystyle  K = \{ x \in {\mathbb R}^n \mid x \geq 0, c^\intercal x = OPT \}.

Notice that this has {n} non-negativity constraints, and one equality constraint. Very simple constraints. Now the oracle asks: given {\alpha \in {\mathbb R}^n, \beta \in {\mathbb R}}, find me a point {x \in K} such that

\displaystyle  \alpha^\intercal x \geq \beta,

or else tell me no such point exists. How would you solve this?

Let’s just try to maximize {\alpha^\intercal x} for {x \in K}. If we can make {\alpha^\intercal x} at least {\beta}, then we can return that point. If we cannot reach {\beta}, we return “No such point.”

OK, suppose all entries of {c} are positive. Then I claim you might as well make {x} non-zero in a single coordinate. Which coordinate? The entry {j} that maximizes

\displaystyle  \frac{OPT}{c_j} \cdot \alpha_j.

Then you’ll make {x_j = \frac{OPT}{c_j}} to satisfy {x \in K}. (Of course, if {\frac{OPT}{c_j} \cdot \alpha_j < \beta}, you’ll return “No such point.”)

If {c} has negative entries, then things are a more interesting. You can still ensure that {x} will be non-zero in at most two entries. And then you have to do a simple case analysis. I’ll omit the details here; let me know if you’d like me to say more.

The next time, we’ll show that if you can solve the oracle problem you can solve LPs (up to {\varepsilon} additive error).

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